Fractional Laplacian

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The fractional Laplacian $(-\Delta)^s$ is a classical operator which gives the standard Laplacian when $s=1$. One can think of $-(-\Delta)^s$ as the most basic elliptic linear integro-differential operator of order $2s$ and can be defined in several equivalent ways (listed below). A range of powers of particular interest is $s \in (0,1)$, in which case for $u \in \mathcal{S}(\mathbb{R}^n)$ we can write the operator as

where $c_{n,s}$ is a universal constant and $\delta u(x,y):= u(x+y)+u(x-y)-2u(x)$. This particular expression shows that in this range of $s$ the operator enjoys the following monotonicity property: if $u$ has a global maximum at $x$, then $(-\Delta)^s u(x) \geq 0$, with equality only if $u$ is constant. From this monotonicity, a comparison principle can be derived for equations involving the fractional Laplacian.

Definitions

All the definitions below are equivalent.

As a pseudo-differential operator

The fractional Laplacian is the pseudo-differential operator with symbol $|\xi|^{2s}$. In other words, the following formula holds
\[ \widehat{(-\Delta)^s f}(\xi) = |\xi|^{2s} \hat f(\xi).\]
for any function (or tempered distribution) for which the right hand side makes sense.

This formula is the simplest to understand and it is useful for problems in the whole space. On the other hand, it is hard to obtain local estimates from it.

From functional calculus

Since the operator $-\Delta$ is a self-adjoint positive definite operator in a dense subset $D$ of $L^2(\R^n)$, one can define $F(-\Delta)$ for any continuous function $F:\R^+ \to \R$. In particular, this serves as a more or less abstract definition of $(-\Delta)^s$.

This definition is not as useful for practical applications, since it does not provide any explicit formula.

The operator can be defined as the generator of $\alpha$-stable Lévy processes. More precisely, if $X_t$ is the isotropic $\alpha$-stable Lévy process starting at zero and $f$ is a smooth function, then
\[ (-\Delta)^{\alpha/2} f(x) = \lim_{h \to 0^+} \frac 1 {h} \mathbb E [f(x) - f(x+X_h)]. \]

This definition is important for applications to probability.

Inverse operator

The inverse of the $s$ power of the Laplacian is the $-s$ power of the Laplacian $(-\Delta)^{-s}$. For $0<s<n/2$, there is an integral formula which says that $(-\Delta)^{-s}u$ is the convolution of the function $u$ with the Riesz potential:
\[ (-\Delta)^{-s} u(x) = C_{n,s} \int_{\R^n} u(x-y) \frac{1}{|y|^{n-2s}} \mathrm d y,\]
which holds as long as $u$ is regular enough for the right hand side to make sense.

The kernel is easy to compute in Fourier side as $\hat p(t,\xi) = e^{-t|\xi|^{2s}}$. There is no explicit formula in physical variables, but the following inequalities are known to hold for some constant $C$
\[ C^{-1} \left( t^{-\frac n {2s}} \wedge \frac{t}{|x|^{n+2s}} \right) \leq p(t,x) \leq C \left( t^{-\frac n {2s}} \wedge \frac{t}{|x|^{n+2s}} \right). \]

Regularity issues

Any function $u$ which satisfies $(-\Delta)^s u=0$ in any open set $\Omega$, then $u \in C^\infty$ inside $\Omega$. This follows from the smoothness of the Poisson kernel for balls.

Full space regularization of the Riesz potential

If $(-\Delta)^s u = f$ in $\R^n$, then of course $u = (-\Delta)^{-s}f$. It is simple to see that the operator $(-\Delta)^{-s}$ regularizes the functions up to $2s$ derivatives. In Fourier side, $\hat u(\xi) = |\xi|^{-2s} \hat f(\xi)$, thus $\hat u$ has a stronger decay than $\hat f$. More precisely, if $f \in C^\alpha$, then $u \in C^{2s+\alpha}$ as long as $2s+\alpha$ is not an integer (A proof of this using only the integral representation of $(-\Delta)^{-s}$ was given in the preliminaries section of [4], but the result is presumably very classical). More generally, if $f$ belongs to the Besov space $B_{p,q}^r$, then $u \in B_{p,q}^{r+2s}$.

Boundary regularity

From the Poisson formula, one can observe that if the boundary data $g$ of the Dirichlet problem in $B_1$ is bounded and smooth, then $u \in C^s(\overline B_1)$ and in general no better. The singularity of $u$ occurs only on $\partial B_1$, the solution $u$ would be $C^\infty$ in the interior of the unit ball (which is also a consequence of the explicit Poisson kernel).